Difference between revisions of "2006 SMT/Calculus Problems/Problem 2"
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| + | ==Problem 2== | ||
| + | Given the equation <math> 4y''+3y'-y=0 </math> and its solution <math> y=e^{\lambda t} </math>, what are the values of <math> \lambda </math>? | ||
| + | |||
==Solution== | ==Solution== | ||
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\frac{0}{e^{\lambda t}}&=\frac{1}{e^{\lambda t}}(4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t})\\ | \frac{0}{e^{\lambda t}}&=\frac{1}{e^{\lambda t}}(4\lambda^{2}e^{\lambda t}+3\lambda e^{\lambda t}-e^{\lambda t})\\ | ||
0&=4\lambda^{2}+3\lambda - 1\\ | 0&=4\lambda^{2}+3\lambda - 1\\ | ||
| − | 0&=(4\lambda-1)(\lambda+1) \Rightarrow \lambda = \frac{1}{4}, 1\\ | + | 0&=(4\lambda-1)(\lambda+1) \Rightarrow \lambda = \frac{1}{4}, -1\\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | Therefore, the possible values of <math>\lambda</math> are <math>\boxed{\lambda = \frac{1}{4}, 1}</math> | + | Therefore, the possible values of <math>\lambda</math> are <math>\boxed{\lambda = \frac{1}{4}, -1}</math> |
Latest revision as of 11:37, 15 January 2020
Problem 2
Given the equation 
and its solution 
, what are the values of 
?
Solution
Substituting into the differential equation, we can solve for the values of :
Dividing through by 
:
Therefore, the possible values of are 
